Integrand size = 21, antiderivative size = 92 \[ \int \frac {e^{\text {csch}^{-1}(c x)} x^5}{1+c^2 x^2} \, dx=-\frac {x}{c^5}-\frac {3 \sqrt {1+\frac {1}{c^2 x^2}} x^2}{8 c^4}+\frac {x^3}{3 c^3}+\frac {\sqrt {1+\frac {1}{c^2 x^2}} x^4}{4 c^2}+\frac {\arctan (c x)}{c^6}+\frac {3 \text {arctanh}\left (\sqrt {1+\frac {1}{c^2 x^2}}\right )}{8 c^6} \]
-x/c^5+1/3*x^3/c^3+arctan(c*x)/c^6+3/8*arctanh((1+1/c^2/x^2)^(1/2))/c^6-3/ 8*x^2*(1+1/c^2/x^2)^(1/2)/c^4+1/4*x^4*(1+1/c^2/x^2)^(1/2)/c^2
Time = 0.09 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.92 \[ \int \frac {e^{\text {csch}^{-1}(c x)} x^5}{1+c^2 x^2} \, dx=\frac {c x \left (-24-9 c \sqrt {1+\frac {1}{c^2 x^2}} x+8 c^2 x^2+6 c^3 \sqrt {1+\frac {1}{c^2 x^2}} x^3\right )+24 \arctan (c x)+9 \log \left (\left (1+\sqrt {1+\frac {1}{c^2 x^2}}\right ) x\right )}{24 c^6} \]
(c*x*(-24 - 9*c*Sqrt[1 + 1/(c^2*x^2)]*x + 8*c^2*x^2 + 6*c^3*Sqrt[1 + 1/(c^ 2*x^2)]*x^3) + 24*ArcTan[c*x] + 9*Log[(1 + Sqrt[1 + 1/(c^2*x^2)])*x])/(24* c^6)
Time = 0.35 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.11, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {6896, 254, 798, 52, 52, 73, 221, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {x^5 e^{\text {csch}^{-1}(c x)}}{c^2 x^2+1} \, dx\) |
\(\Big \downarrow \) 6896 |
\(\displaystyle \frac {\int \frac {x^4}{c^2 x^2+1}dx}{c}+\frac {\int \frac {x^3}{\sqrt {1+\frac {1}{c^2 x^2}}}dx}{c^2}\) |
\(\Big \downarrow \) 254 |
\(\displaystyle \frac {\int \frac {x^3}{\sqrt {1+\frac {1}{c^2 x^2}}}dx}{c^2}+\frac {\int \left (\frac {x^2}{c^2}+\frac {1}{c^4 \left (c^2 x^2+1\right )}-\frac {1}{c^4}\right )dx}{c}\) |
\(\Big \downarrow \) 798 |
\(\displaystyle \frac {\int \left (\frac {x^2}{c^2}+\frac {1}{c^4 \left (c^2 x^2+1\right )}-\frac {1}{c^4}\right )dx}{c}-\frac {\int \frac {x^6}{\sqrt {1+\frac {1}{c^2 x^2}}}d\frac {1}{x^2}}{2 c^2}\) |
\(\Big \downarrow \) 52 |
\(\displaystyle \frac {\int \left (\frac {x^2}{c^2}+\frac {1}{c^4 \left (c^2 x^2+1\right )}-\frac {1}{c^4}\right )dx}{c}-\frac {-\frac {3 \int \frac {x^4}{\sqrt {1+\frac {1}{c^2 x^2}}}d\frac {1}{x^2}}{4 c^2}-\frac {1}{2} x^4 \sqrt {\frac {1}{c^2 x^2}+1}}{2 c^2}\) |
\(\Big \downarrow \) 52 |
\(\displaystyle \frac {\int \left (\frac {x^2}{c^2}+\frac {1}{c^4 \left (c^2 x^2+1\right )}-\frac {1}{c^4}\right )dx}{c}-\frac {-\frac {3 \left (x^2 \left (-\sqrt {\frac {1}{c^2 x^2}+1}\right )-\frac {\int \frac {x^2}{\sqrt {1+\frac {1}{c^2 x^2}}}d\frac {1}{x^2}}{2 c^2}\right )}{4 c^2}-\frac {1}{2} x^4 \sqrt {\frac {1}{c^2 x^2}+1}}{2 c^2}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {\int \left (\frac {x^2}{c^2}+\frac {1}{c^4 \left (c^2 x^2+1\right )}-\frac {1}{c^4}\right )dx}{c}-\frac {-\frac {3 \left (x^2 \left (-\sqrt {\frac {1}{c^2 x^2}+1}\right )-\int \frac {1}{\frac {c^2}{x^4}-c^2}d\sqrt {1+\frac {1}{c^2 x^2}}\right )}{4 c^2}-\frac {1}{2} x^4 \sqrt {\frac {1}{c^2 x^2}+1}}{2 c^2}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {\int \left (\frac {x^2}{c^2}+\frac {1}{c^4 \left (c^2 x^2+1\right )}-\frac {1}{c^4}\right )dx}{c}-\frac {-\frac {3 \left (\frac {\text {arctanh}\left (\sqrt {\frac {1}{c^2 x^2}+1}\right )}{c^2}-x^2 \sqrt {\frac {1}{c^2 x^2}+1}\right )}{4 c^2}-\frac {1}{2} x^4 \sqrt {\frac {1}{c^2 x^2}+1}}{2 c^2}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {\arctan (c x)}{c^5}-\frac {x}{c^4}+\frac {x^3}{3 c^2}}{c}-\frac {-\frac {3 \left (\frac {\text {arctanh}\left (\sqrt {\frac {1}{c^2 x^2}+1}\right )}{c^2}-x^2 \sqrt {\frac {1}{c^2 x^2}+1}\right )}{4 c^2}-\frac {1}{2} x^4 \sqrt {\frac {1}{c^2 x^2}+1}}{2 c^2}\) |
(-(x/c^4) + x^3/(3*c^2) + ArcTan[c*x]/c^5)/c - (-1/2*(Sqrt[1 + 1/(c^2*x^2) ]*x^4) - (3*(-(Sqrt[1 + 1/(c^2*x^2)]*x^2) + ArcTanh[Sqrt[1 + 1/(c^2*x^2)]] /c^2))/(4*c^2))/(2*c^2)
3.1.60.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( m + n + 2)/((b*c - a*d)*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && ILtQ[m, -1] && FractionQ[n] && LtQ[n, 0]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[(x_)^(m_)/((a_) + (b_.)*(x_)^2), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^2, x], x] /; FreeQ[{a, b}, x] && IGtQ[m, 3]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[1/n Subst [Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]
Int[(E^ArcCsch[(c_.)*(x_)]*((d_.)*(x_))^(m_.))/((a_) + (b_.)*(x_)^2), x_Sym bol] :> Simp[d^2/(a*c^2) Int[(d*x)^(m - 2)/Sqrt[1 + 1/(c^2*x^2)], x], x] + Simp[d/c Int[(d*x)^(m - 1)/(a + b*x^2), x], x] /; FreeQ[{a, b, c, d, m} , x] && EqQ[b - a*c^2, 0]
Leaf count of result is larger than twice the leaf count of optimal. \(171\) vs. \(2(78)=156\).
Time = 0.96 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.87
method | result | size |
default | \(\frac {\sqrt {\frac {c^{2} x^{2}+1}{c^{2} x^{2}}}\, x \left (2 x \left (\frac {c^{2} x^{2}+1}{c^{2}}\right )^{\frac {3}{2}} c^{4}-5 x \sqrt {\frac {c^{2} x^{2}+1}{c^{2}}}\, c^{2}-5 \ln \left (x +\sqrt {\frac {c^{2} x^{2}+1}{c^{2}}}\right )+8 \ln \left (x +\sqrt {-\frac {\left (-c^{2} x +\sqrt {-c^{2}}\right ) \left (c^{2} x +\sqrt {-c^{2}}\right )}{c^{4}}}\right )\right )}{8 \sqrt {\frac {c^{2} x^{2}+1}{c^{2}}}\, c^{6}}+\frac {\frac {\frac {1}{3} c^{2} x^{3}-x}{c^{4}}+\frac {\arctan \left (c x \right )}{c^{5}}}{c}\) | \(172\) |
1/8*((c^2*x^2+1)/c^2/x^2)^(1/2)*x*(2*x*(1/c^2*(c^2*x^2+1))^(3/2)*c^4-5*x*( 1/c^2*(c^2*x^2+1))^(1/2)*c^2-5*ln(x+(1/c^2*(c^2*x^2+1))^(1/2))+8*ln(x+(-(- c^2*x+(-c^2)^(1/2))*(c^2*x+(-c^2)^(1/2))/c^4)^(1/2)))/(1/c^2*(c^2*x^2+1))^ (1/2)/c^6+1/c*(1/c^4*(1/3*c^2*x^3-x)+1/c^5*arctan(c*x))
Time = 0.27 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.98 \[ \int \frac {e^{\text {csch}^{-1}(c x)} x^5}{1+c^2 x^2} \, dx=\frac {8 \, c^{3} x^{3} - 24 \, c x + 3 \, {\left (2 \, c^{4} x^{4} - 3 \, c^{2} x^{2}\right )} \sqrt {\frac {c^{2} x^{2} + 1}{c^{2} x^{2}}} + 24 \, \arctan \left (c x\right ) - 9 \, \log \left (c x \sqrt {\frac {c^{2} x^{2} + 1}{c^{2} x^{2}}} - c x\right )}{24 \, c^{6}} \]
1/24*(8*c^3*x^3 - 24*c*x + 3*(2*c^4*x^4 - 3*c^2*x^2)*sqrt((c^2*x^2 + 1)/(c ^2*x^2)) + 24*arctan(c*x) - 9*log(c*x*sqrt((c^2*x^2 + 1)/(c^2*x^2)) - c*x) )/c^6
\[ \int \frac {e^{\text {csch}^{-1}(c x)} x^5}{1+c^2 x^2} \, dx=\frac {\int \frac {x^{4}}{c^{2} x^{2} + 1}\, dx + \int \frac {c x^{5} \sqrt {1 + \frac {1}{c^{2} x^{2}}}}{c^{2} x^{2} + 1}\, dx}{c} \]
(Integral(x**4/(c**2*x**2 + 1), x) + Integral(c*x**5*sqrt(1 + 1/(c**2*x**2 ))/(c**2*x**2 + 1), x))/c
Leaf count of result is larger than twice the leaf count of optimal. 162 vs. \(2 (78) = 156\).
Time = 0.31 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.76 \[ \int \frac {e^{\text {csch}^{-1}(c x)} x^5}{1+c^2 x^2} \, dx=\frac {c^{2} x^{3} - 3 \, x}{3 \, c^{5}} - \frac {\frac {2 \, {\left (\frac {5 \, \sqrt {\frac {c^{2} x^{2} + 1}{x^{2}}}}{c} - \frac {3 \, \left (\frac {c^{2} x^{2} + 1}{x^{2}}\right )^{\frac {3}{2}}}{c^{3}}\right )}}{\frac {2 \, {\left (c^{2} x^{2} + 1\right )}}{c^{2} x^{2}} - \frac {{\left (c^{2} x^{2} + 1\right )}^{2}}{c^{4} x^{4}} - 1} - 3 \, \log \left (\frac {\sqrt {\frac {c^{2} x^{2} + 1}{x^{2}}}}{c} + 1\right ) + 3 \, \log \left (\frac {\sqrt {\frac {c^{2} x^{2} + 1}{x^{2}}}}{c} - 1\right )}{16 \, c^{6}} + \frac {\arctan \left (c x\right )}{c^{6}} \]
1/3*(c^2*x^3 - 3*x)/c^5 - 1/16*(2*(5*sqrt((c^2*x^2 + 1)/x^2)/c - 3*((c^2*x ^2 + 1)/x^2)^(3/2)/c^3)/(2*(c^2*x^2 + 1)/(c^2*x^2) - (c^2*x^2 + 1)^2/(c^4* x^4) - 1) - 3*log(sqrt((c^2*x^2 + 1)/x^2)/c + 1) + 3*log(sqrt((c^2*x^2 + 1 )/x^2)/c - 1))/c^6 + arctan(c*x)/c^6
Time = 0.28 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.97 \[ \int \frac {e^{\text {csch}^{-1}(c x)} x^5}{1+c^2 x^2} \, dx=\frac {1}{8} \, \sqrt {c^{2} x^{2} + 1} x {\left (\frac {2 \, x^{2} {\left | c \right |} \mathrm {sgn}\left (x\right )}{c^{4}} - \frac {3 \, {\left | c \right |} \mathrm {sgn}\left (x\right )}{c^{6}}\right )} - \frac {3 \, \log \left (-x {\left | c \right |} + \sqrt {c^{2} x^{2} + 1}\right ) \mathrm {sgn}\left (x\right )}{8 \, c^{6}} + \frac {\arctan \left (c x\right )}{c^{6}} + \frac {c^{6} x^{3} - 3 \, c^{4} x}{3 \, c^{9}} \]
1/8*sqrt(c^2*x^2 + 1)*x*(2*x^2*abs(c)*sgn(x)/c^4 - 3*abs(c)*sgn(x)/c^6) - 3/8*log(-x*abs(c) + sqrt(c^2*x^2 + 1))*sgn(x)/c^6 + arctan(c*x)/c^6 + 1/3* (c^6*x^3 - 3*c^4*x)/c^9
Time = 5.34 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.86 \[ \int \frac {e^{\text {csch}^{-1}(c x)} x^5}{1+c^2 x^2} \, dx=\frac {3\,\mathrm {atanh}\left (\sqrt {\frac {1}{c^2\,x^2}+1}\right )}{8\,c^6}+\frac {3\,\mathrm {atan}\left (c\,x\right )-3\,c\,x+c^3\,x^3}{3\,c^6}+\frac {x^4\,\sqrt {\frac {1}{c^2\,x^2}+1}}{4\,c^2}-\frac {3\,x^2\,\sqrt {\frac {1}{c^2\,x^2}+1}}{8\,c^4} \]